krit.club logo

Laws of Motion - Newton's First, Second, and Third Laws

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

β€’

Newton's First Law (Law of Inertia): A body continues to be in its state of rest or of uniform motion in a straight line unless compelled by an external force to change that state. Inertia is the inherent property of an object to resist change, and its measure is mass mm.

β€’

Linear Momentum: The linear momentum of a body is defined as the product of its mass mm and velocity v⃗\vec{v}. It is a vector quantity denoted by p⃗=mv⃗\vec{p} = m\vec{v}. Its SI unit is kg⋅m/skg \cdot m/s.

β€’

Newton's Second Law: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Mathematically, Fβƒ—βˆdpβƒ—dt\vec{F} \propto \frac{d\vec{p}}{dt}.

β€’

Impulse: When a large force acts on a body for a short interval of time, the product of force and time is called Impulse. Impulse is equal to the total change in momentum: J⃗=F⃗avgΔt=Δp⃗\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}.

β€’

Newton's Third Law: To every action, there is always an equal and opposite reaction. If body AA exerts a force Fβƒ—AB\vec{F}_{AB} on body BB, then body BB exerts a force Fβƒ—BA\vec{F}_{BA} on body AA such that Fβƒ—AB=βˆ’Fβƒ—BA\vec{F}_{AB} = -\vec{F}_{BA}.

β€’

Equilibrium of a Particle: A particle is in equilibrium when the vector sum of all external forces acting on it is zero, i.e., βˆ‘Fβƒ—=0\sum \vec{F} = 0.

πŸ“Formulae

p⃗=mv⃗\vec{p} = m\vec{v}

F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}

F⃗=ma⃗ (when mass is constant)\vec{F} = m\vec{a} \text{ (when mass is constant)}

Jβƒ—=∫Fβƒ—dt=Ξ”pβƒ—=m(vβƒ—βˆ’uβƒ—)\vec{J} = \int \vec{F} dt = \Delta \vec{p} = m(\vec{v} - \vec{u})

Fβƒ—AB=βˆ’Fβƒ—BA\vec{F}_{AB} = -\vec{F}_{BA}

πŸ’‘Examples

Problem 1:

A constant force acting on a body of mass 3.0kg3.0 kg changes its speed from 2.0m/s2.0 m/s to 3.5m/s3.5 m/s in 25s25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Solution:

Given: m=3.0kgm = 3.0 kg, u=2.0m/su = 2.0 m/s, v=3.5m/sv = 3.5 m/s, t=25st = 25 s. Using the formula F=ma=mvβˆ’utF = ma = m\frac{v - u}{t}, we get F=3.0Γ—3.5βˆ’2.025=3.0Γ—1.525=0.18NF = 3.0 \times \frac{3.5 - 2.0}{25} = 3.0 \times \frac{1.5}{25} = 0.18 N.

Explanation:

Since the speed increases in the same direction, the force acts in the direction of motion with a magnitude of 0.18N0.18 N.

Problem 2:

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12m/s12 m/s. If the mass of the ball is 0.15kg0.15 kg, determine the impulse imparted to the ball.

Solution:

Given: m=0.15kgm = 0.15 kg, u=12m/su = 12 m/s, v=βˆ’12m/sv = -12 m/s (opposite direction). Impulse J=Ξ”p=m(vβˆ’u)=0.15Γ—(βˆ’12βˆ’12)=0.15Γ—(βˆ’24)=βˆ’3.6NsJ = \Delta p = m(v - u) = 0.15 \times (-12 - 12) = 0.15 \times (-24) = -3.6 Ns.

Explanation:

The impulse is 3.6Ns3.6 Ns directed towards the bowler. The negative sign indicates the direction is opposite to the initial velocity.

Newton's First, Second, and Third Laws Revision - Class 11 Physics CBSE